Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# NCERT Solutions for Class 7 Maths Chapter 10: Algebraic Expressions - Exercise 10.2

Last updated date: 10th Sep 2024
Total views: 612.6k
Views today: 13.12k

## NCERT Solutions for Class 7 Maths Chapter 10 (EX 10.2)

Free PDF download of NCERT Solutions for Class 7 Maths Chapter 10 Exercise 10.2 (EX 10.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 7 Maths Chapter 10 Algebraic Expressions Exercise 10.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

 Class: NCERT Solutions for Class 7 Subject: Class 7 Maths Chapter Name: Chapter 10 - Algebraic Expressions Exercise: Exercise - 10.2 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes

Download NCERT Solutions PDF and opt to cross-refer post-answering questions to score subject-best marks. Subjects like Science, Maths, English, Social Science, Hindi will become easy to study if you have access to NCERT Solution for Class 7 Science , Maths solutions and solutions of other subjects.

## Access NCERT Solutions for Class 7 Maths Chapter 10 - Algebraic Expressions

### Exercise 10.2

1. If $\mathbf{m = 2}$, find the value of:

(i)$\mathbf{m - 2}$

Ans: Substitute $2$ for $m$ and solve.

$\Rightarrow m - 2 = 2 - 2 \\$

$\Rightarrow m - 2 = 0 \\$

The value is $0$.

(ii) $\mathbf{3m - 5}$

Ans: Substitute $2$ for $m$ and solve.

$\Rightarrow 3m - 5 = 3\left( 2 \right) - 5 \\$

$\Rightarrow 3m - 5 = 6 - 5 \\$

$\Rightarrow 3m - 5 = 1 \\$

The value is $1$.

(iii) $\mathbf{9 - 5m}$

Ans: Substitute $2$ for $m$ and solve.

$\Rightarrow 9 - 5m = 9 - 5\left( 2 \right) \\$

$\Rightarrow 9 - 5m = 9 - 10 \\$

$\Rightarrow 9 - 5m = - 1 \\$

The value is $- 1$.

(iv)$\mathbf{3{m^2} - 2m - 7}$

Ans: Substitute $2$ for $m$ and solve.

$\Rightarrow 3{m^2} - 2m - 7 = 3{\left( 2 \right)^2} - 2\left( 2 \right) - 7 \\$

$\Rightarrow 3{m^2} - 2m - 7 = 3\left( 4 \right) - 4 - 7 \\$

$\Rightarrow 3{m^2} - 2m - 7 = 12 - 11 \\$

$\Rightarrow 3{m^2} - 2m - 7 = 1 \\$

The value is $1$.

(v) $\mathbf{\dfrac{{5m}}{2} - 4}$

Ans: Substitute $2$ for $m$ and solve.

$\Rightarrow \dfrac{{5m}}{2} - 4 = \dfrac{{5 \times 2}}{2} - 4 \\$

$\Rightarrow \dfrac{{5m}}{2} - 4 = 5 - 4 \\$

$\Rightarrow \dfrac{{5m}}{2} - 4 = 1 \\$

The value is $1$.

2. If $\mathbf{p = - 2}$, find the value of:

(i) $\mathbf{4p + 7}$

Ans: Substitute $- 2$ for $p$ and solve.

$\Rightarrow 4p + 7 = 4\left( { - 2} \right) + 7 \\$

$\Rightarrow 4p + 7 = - 8 + 7 \\$

$\Rightarrow 4p + 7 = - 1 \\$

The value is $- 1$.

(ii) $\mathbf{ - 3{p^2} + 4p + 7}$

Ans: Substitute $- 2$ for $p$ and solve.

$\Rightarrow - 3{p^2} + 4p + 7 = - 3{\left( { - 2} \right)^2} + 4\left( { - 2} \right) + 7 \\$

$\Rightarrow - 3{p^2} + 4p + 7 = - 3\left( 4 \right) - 8 + 7 \\$

$\Rightarrow - 3{p^2} + 4p + 7 = - 12 - 1 \\$

$\Rightarrow - 3{p^2} + 4p + 7 = - 13 \\$

The value is $- 13$.

(iii) $\mathbf{ - 2{p^3} - 3{p^2} + 4p + 7}$

Ans: Substitute $- 2$ for $p$ and solve.

$\Rightarrow - 2{p^3} - 3{p^2} + 4p + 7 = - 2{\left( { - 2} \right)^3} - 3{\left( { - 2} \right)^2} + 4\left( { - 2} \right) + 7 \\$

$\Rightarrow - 2{p^3} - 3{p^2} + 4p + 7 = - 2 \times \left( { - 8} \right) - 3\left( 4 \right) - 8 + 7 \\$

$\Rightarrow - 2{p^3} - 3{p^2} + 4p + 7 = 16 - 12 - 1 \\$

$\Rightarrow - 2{p^3} - 3{p^2} + 4p + 7 = 4 - 1 \\$

$\Rightarrow - 2{p^3} - 3{p^2} + 4p + 7 = 3 \\$

The value is $3$.

3. Find the value of the following expressions, when $\mathbf{x = - 1}$:

(i) $\mathbf{2x - 7}$

Ans: Substitute $- 1$ for $x$ and solve.

$\Rightarrow 2x - 7 = 2\left( { - 1} \right) - 7 \\$

$\Rightarrow 2x - 7 = - 2 - 7 \\$

$\Rightarrow 2x - 7 = - 9 \\$

The value of expression $2x - 7$ is $- 9$.

(ii) $\mathbf{- x + 2}$

Ans: Substitute $- 1$ for $x$ and solve.

$\Rightarrow - x + 2 = - \left( { - 1} \right) + 2 \\$

$\Rightarrow - x + 2 = 1 + 2 \\$

$\Rightarrow - x + 2 = 3 \\$

The value of expression $- x + 2$ is $3$.

(iii) $\mathbf{{x^2} + 2x + 1}$

Ans: Substitute $- 1$ for $x$ and solve.

$\Rightarrow {x^2} + 2x + 1 = {\left( { - 1} \right)^2} + 2\left( { - 1} \right) + 1 \\$

$\Rightarrow {x^2} + 2x + 1 = 1 - 2 + 1 \\$

$\Rightarrow {x^2} + 2x + 1 = - 1 + 1 \\$

$\Rightarrow {x^2} + 2x + 1 = 0 \\$

The value of expression ${x^2} + 2x + 1$ is $0$.

(iv)$\mathbf{2{x^2} - x - 2}$

Ans: Substitute $- 1$ for $x$ and solve.

$\Rightarrow 2{x^2} - x - 2 = 2{\left( { - 1} \right)^2} - \left( { - 1} \right) - 2 \\$

$\Rightarrow 2{x^2} - x - 2 = 2 \times 1 + 1 - 2 \\$

$\Rightarrow 2{x^2} - x - 2 = 2 + 1 - 2 \\$

$\Rightarrow 2{x^2} - x - 2 = 3 - 2 \\$

$\Rightarrow 2{x^2} - x - 2 = 1 \\$

The value of expression $2{x^2} - x - 2$ is $1$.

4. If $\mathbf{a = 2$, $b = - 2}$, find the value of:

(i) $\mathbf{{a^2} + {b^2}}$

Ans: Substitute $2$ for $a$ and $- 2$ for $b$ then solve.

$\Rightarrow {a^2} + {b^2} = {\left( 2 \right)^2} + {\left( { - 2} \right)^2} \\$

$\Rightarrow {a^2} + {b^2} = 4 + 4 \\$

$\Rightarrow {a^2} + {b^2} = 8 \\$

The value is $8$.

(ii) ${a^2} + ab + {b^2}$

Ans: Substitute $2$ for $a$ and $- 2$ for $b$ then solve.

$\Rightarrow {a^2} + ab + {b^2} = {\left( 2 \right)^2} + \left( 2 \right)\left( { - 2} \right) + {\left( { - 2} \right)^2} \\$

$\Rightarrow {a^2} + ab + {b^2} = 4 - 4 + 4 \\$

$\Rightarrow {a^2} + ab + {b^2} = 4 \\$

The value is $4$.

(iii) ${a^2} - {b^2}$

Ans: Substitute $2$ for $a$ and $- 2$ for $b$ then solve.

$\Rightarrow {a^2} - {b^2} = {\left( 2 \right)^2} - {\left( { - 2} \right)^2} \\$

$\Rightarrow {a^2} - {b^2} = 4 - 4 \\$

$\Rightarrow {a^2} - {b^2} = 0 \\$

The value is $0$.

5. When $\mathbf{a = 0$, $b = - 1}$, find the value of the given expressions:

(i) $\mathbf{2a + 2b}$

Ans: Substitute $0$ for $a$ and $- 1$ for $b$ then solve.

$\Rightarrow 2a + 2b = 2\left( 0 \right) + 2\left( { - 1} \right) \\$

$\Rightarrow 2a + 2b = 0 - 2 \\$

$\Rightarrow 2a + 2b = - 2 \\$

The value of expression $2a + 2b$ is $- 2$.

(ii) $\mathbf{2{a^2} + {b^2} + 1}$

Ans: Substitute $0$ for $a$ and $- 1$ for $b$ then solve.

$\Rightarrow 2{a^2} + {b^2} + 1 = 2{\left( 0 \right)^2} + {\left( { - 1} \right)^2} + 1 \\$

$\Rightarrow 2{a^2} + {b^2} + 1 = 2 \times 0 + 1 + 1 \\$

$\Rightarrow 2{a^2} + {b^2} + 1 = 0 + 2 \\$

$\Rightarrow 2{a^2} + {b^2} + 1 = 2 \\$

The value of expression $2{a^2} + {b^2} + 1$ is $2$.

(iii) $\mathbf{2{a^2}b + 2a{b^2} + ab}$

Ans: Substitute $0$ for $a$ and $- 1$ for $b$ then solve.

$\Rightarrow 2{a^2}b + 2a{b^2} + ab = 2{\left( 0 \right)^2}\left( { - 1} \right) + 2\left( 0 \right){\left( { - 1} \right)^2} + \left( 0 \right)\left( { - 1} \right) \\$

$\Rightarrow 2{a^2}b + 2a{b^2} + ab = 0 + 0 + 0 \\$

$\Rightarrow 2{a^2}b + 2a{b^2} + ab = 0 \\$

The value of expression $2{a^2}b + 2a{b^2} + ab$ is $0$.

(iv)$\mathbf{{a^2} + ab + 2}$

Ans: Substitute $0$ for $a$ and $- 1$ for $b$ then solve.

$\Rightarrow {a^2} + ab + 2 = {\left( 0 \right)^2} + \left( 0 \right)\left( { - 1} \right) + 2 \\$

$\Rightarrow {a^2} + ab + 2 = 0 + 0 + 2 \\$

$\Rightarrow {a^2} + ab + 2 = 2 \\$

The value of expression ${a^2} + ab + 2$ is $2$.

6. Simplify the expressions and find the value if $\mathbf{x}$ is equal to $\mathbf{2}$:

(i) $\mathbf{x + 7 + 4\left( {x - 5} \right)}$

Ans: Simplify the brackets and combine the like terms.

$\Rightarrow x + 7 + 4\left( {x - 5} \right) = x + 7 + 4x - 20 \\$

$\Rightarrow x + 7 + 4\left( {x - 5} \right) = x + 4x + 7 - 20 \\$

$\Rightarrow x + 7 + 4\left( {x - 5} \right) = 5x - 13 \\$

Substitute $2$ for $x$ and then solve.

$\Rightarrow x + 7 + 4\left( {x - 5} \right) = 5\left( 2 \right) - 13 \\$

$\Rightarrow x + 7 + 4\left( {x - 5} \right) = 10 - 13 \\$

$\Rightarrow x + 7 + 4\left( {x - 5} \right) = - 3 \\$

The determined value is $- 3$.

(ii) $\mathbf{3\left( {x + 2} \right) + 5x - 7}$

Ans: Simplify the brackets and combine the like terms.

$\Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 3x + 6 + 5x - 7 \\$

$\Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 3x + 5x + 6 - 7 \\$

$\Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 8x - 1 \\$

Substitute $2$ for $x$ and then solve.

$\Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 8\left( 2 \right) - 1 \\$

$\Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 16 - 1 \\$

$\Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 15 \\$

The determined value is $15$.

(iii) $\mathbf{6x + 5\left( {x - 2} \right)}$

Ans: Simplify the brackets and combine the like terms.

$\Rightarrow 6x + 5\left( {x - 2} \right) = 6x + 5x - 10 \\$

$\Rightarrow 6x + 5\left( {x - 2} \right) = 11x - 10 \\$

Substitute $2$ for $x$ and then solve.

$\Rightarrow 6x + 5\left( {x - 2} \right) = 11 \times 2 - 10 \\$

$\Rightarrow 6x + 5\left( {x - 2} \right) = 22 - 10 \\$

$\Rightarrow 6x + 5\left( {x - 2} \right) = 12 \\$

The determined value is $12$.

(iv)$\mathbf{4\left( {2x - 1} \right) + 3x + 11}$

Ans: Simplify the brackets and combine the like terms.

$\Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 8x - 4 + 3x + 11 \\$

$\Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 8x + 3x - 4 + 11 \\$

$\Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 11x + 7 \\$

Substitute $2$ for $x$ and then solve.

$\Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 11 \times 2 + 7 \\$

$\Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 22 + 7 \\$

$\Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 29 \\$

The determined value is $29$.

7. Simplify these expressions and find their values if $\mathbf{x = 3$, $a = - 1$, $b = - 2}$:

(i) $\mathbf{3x - 5 - x + 9}$

Ans: Simplify the brackets and combine the like terms.

$\Rightarrow 3x - 5 - x + 9 = 3x - x - 5 + 9 \\$

$\Rightarrow 3x - 5 - x + 9 = 2x + 4 \\$

Substitute $3$ for $x$ and then solve.

$\Rightarrow 3x - 5 - x + 9 = 2 \times 3 + 4 \\$

$\Rightarrow 3x - 5 - x + 9 = 6 + 4 \\$

$\Rightarrow 3x - 5 - x + 9 = 10 \\$

The determined value is $10$.

(ii) $\mathbf{2 - 8x + 4x + 4}$

Ans: Simplify the brackets and combine the like terms.

$\Rightarrow 2 - 8x + 4x + 4 = - 8x + 4x + 2 + 4 \\$

$\Rightarrow 2 - 8x + 4x + 4 = - 4x + 6 \\$

Substitute $3$ for $x$ and then solve.

$\Rightarrow 2 - 8x + 4x + 4 = - 4 \times 3 + 6 \\$

$\Rightarrow 2 - 8x + 4x + 4 = - 12 + 6 \\$

$\Rightarrow 2 - 8x + 4x + 4 = - 6 \\$

The determined value is $- 6$.

(iii) $\mathbf{3a + 5 - 8a + 1}$

Ans: Simplify the brackets and combine the like terms.

$\Rightarrow 3a + 5 - 8a + 1 = 3a - 8a + 5 + 1 \\$

$\Rightarrow 3a + 5 - 8a + 1 = - 5a + 6 \\$

Substitute $- 1$ for $a$ and then solve.

$\Rightarrow 3a + 5 - 8a + 1 = - 5\left( { - 1} \right) + 6 \\$

$\Rightarrow 3a + 5 - 8a + 1 = 5 + 6 \\$

$\Rightarrow 3a + 5 - 8a + 1 = 11 \\$

The determined value is $11$.

(iv)$\mathbf{10 - 3b - 4 - 5b}$

Ans: Simplify the brackets and combine the like terms.

$\Rightarrow 10 - 3b - 4 - 5b = - 3b - 5b + 10 - 4 \\$

$\Rightarrow 10 - 3b - 4 - 5b = - 8b + 6 \\$

Substitute $- 2$ for $b$ and then solve.

$\Rightarrow 10 - 3b - 4 - 5b = - 8\left( { - 2} \right) + 6 \\$

$\Rightarrow 10 - 3b - 4 - 5b = 16 + 6 \\$

$\Rightarrow 10 - 3b - 4 - 5b = 22 \\$

The determined value is $22$.

(v) $\mathbf{2a - 2b - 4 - 5 + a}$

Ans: Simplify the brackets and combine the like terms.

$\Rightarrow 2a - 2b - 4 - 5 + a = 2a + a - 2b - 4 - 5 \\$

$\Rightarrow 2a - 2b - 4 - 5 + a = 3a - 2b - 9 \\$

Substitute $- 2$ for $b$ and $- 1$ for $a$ and then solve.

$\Rightarrow 2a - 2b - 4 - 5 + a = 3\left( { - 1} \right) - 2\left( { - 2} \right) - 9 \\$

$\Rightarrow 2a - 2b - 4 - 5 + a = - 3 + 4 - 9 \\$

$\Rightarrow 2a - 2b - 4 - 5 + a = 1 - 9 \\$

$\Rightarrow 2a - 2b - 4 - 5 + a = - 8 \\$

The determined value is $- 8$.

8. (i) If $\mathbf{z = 10}$, find the value of $\mathbf{{z^3} - 3\left( {z - 10} \right)}$.

Ans: Substitute the value of $z$ as $10$ then find a cube of $10$ to solve.

$\Rightarrow {z^3} - 3\left( {z - 10} \right) = {\left( {10} \right)^3} - 3\left( {10 - 10} \right) \\$

$\Rightarrow {z^3} - 3\left( {z - 10} \right) = 1000 - 3 \times 0 \\$

$\Rightarrow {z^3} - 3\left( {z - 10} \right) = 1000 \\$

The value of ${z^3} - 3\left( {z - 10} \right)$ is $1000$.

8. (ii) If $\mathbf{p = - 10}$, find the value of $\mathbf{{p^2} - 2p - 100}$

Ans: Substitute the value of $p$ as $- 10$ then find a square of $10$ to solve.

$\Rightarrow {p^2} - 2p - 100 = {\left( {10} \right)^2} - 2\left( { - 10} \right) - 100 \\$

$\Rightarrow {p^2} - 2p - 100 = 100 + 20 - 100 \\$

$\Rightarrow {p^2} - 2p - 100 = 20 \\$

The value of ${p^2} - 2p - 100$ is $20$.

9. What should be the value of $a$ if the value of $\mathbf{2{x^2} + x - a}$ equals to $5$, when $\mathbf{x = 0}$?

Ans: Substitute $0$ for $x$ in the expression $2{x^2} + x - a$ and equate it to $5$ in order to solve for $a$.

$\Rightarrow 2{x^2} + x - a = 5 \\$

$\Rightarrow 2{\left( 0 \right)^2} + 0 - a = 5 \\$

$\Rightarrow 0 + 0 - a = 5 \\$

$\Rightarrow - a = 5 \\$

$\Rightarrow a = - 5 \\$

So, the value of $a$ is $- 5$.

10. Simplify the expression and find its value when $\mathbf{a = 5}$ and $\mathbf{b = - 3}$:

$\mathbf{2\left( {{a^2} + ab} \right) + 3 - ab}$

Ans: Solve the brackets and combine the like terms with the variable $ab$.

$\Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 2{a^2} + 2ab + 3 - ab \\$

$\Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 2{a^2} + 2ab - ab + 3 \\$

$\Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 2{a^2} + ab + 3 \\$

Now substitute $5$ for $a$ and $- 3$ for $b$ to solve.

$\Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 2{\left( 5 \right)^2} + \left( 5 \right)\left( { - 3} \right) + 3 \\$

$\Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 2 \times 25 - 15 + 3 \\$

$\Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 50 - 12 \\$

$\Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 38 \\$

The value of expression $2\left( {{a^2} + ab} \right) + 3 - ab$ is $38$.

### NCERT Solutions for Class 7 Maths Chapter 10 Algebraic Expressions Exercise 10.2

Opting for the NCERT solutions for Ex 10.2 Class 7 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 10.2 Class 7 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 7 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 7 Maths Chapter 10 Exercise 10.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 7 Maths Chapter 10 Exercise 10.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 7 Maths Chapter 10 Exercise 10.2 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

## FAQs on NCERT Solutions for Class 7 Maths Chapter 10: Algebraic Expressions - Exercise 10.2

1. Is chapter 10 Practical Geometry in NCERT Class 7 Maths important?

Absolutely yes chapter 10 Practical Geometry is a very important chapter in Class 7 Maths. This will give you a base for the upcoming classes and also be useful for any competitive exams. So, students should give this chapter extra focus and understand the concepts. You can refer to Vedantu, where you can get detailed and stepwise solutions to Chapter 10 Practical Geometry in the free PDF format. Also, you will be able to build a better understanding level and learn a different approach to solving the problem of Geometry.

2. How many questions are there in NCERT Solutions for  Class 7 Maths Chapter 10 Practical Geometry (EX 10.2) Exercise 10.2?

There are a total of 4 questions in  Class 7 Maths Chapter 10 Practical Geometry (EX 10.2) Exercise 10.2  Solutions to all these questions are being provided by Vedantu, and these are available on the official website of Vedantu and the mobile application of Vedantu. The best part is that these NCERT solutions were specially designed by qualified teacher experts, and they are available to students free of charge.

3. How beneficial are NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry (EX 10.2) Exercise 10.2 for board exams?

NCERT Solutions for class 7 Maths Chapter 10 Practical Geometry (EX 10.2) Exercise 10.2 provides answers with detailed descriptions as per the syllabus prescribed by the CBSE board. For the students to finish the assignment on time, solving these solutions would be fantastic practice. The NCERT Solutions for class 7 Maths Chapter 10 Practical Geometry (EX 10.2) Exercise 10.2  are obviously necessary to achieve excellent exam scores. Students can practice writing exams and become more comfortable with the process

4. How can NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry (EX 10.2) Exercise 10.2 be helpful in the final exams?

Chapter 10 of Class 7 Math's "Lines and Angles" covers topics like "Triangles, "Angle " etc. Students may have trouble with fundamental issues when working through these NCERT practice solutions. Vedantu offers free NCERT solutions that use a simple methodology so that students can grasp the fundamentals while answering these questions. In addition, they will be ready for challenging questions before their final exams.

5. What are the topics discussed in Class 7 Chapter 6 The Triangle and its Properties?

The topics discussed in Class 7 Chapter 6 The Triangle and Its Properties, are: